Question 5 of 5: Transient Drawdown Near a River (Constant-Head) and Near an Impermeable Boundary

Question 5: Transient Drawdown Near a River (Constant-Head) and Near an Impermeable Boundary (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Confined aquifer, thickness $b=30$ m, $S_s=2.3\times10^{-5}\ \text{m}^{-1}$, $K=1.2\times10^{-5}$ m/s. Pumping rate $Q=50\ \text{m}^3/\text{hr}=0.01389\ \text{m}^3/\text{s}$. Pumping well 200 m west of a linear boundary. (a) Constant-head (river) boundary; observation well 60 m south of the pumping well; $t=24$ and 36 h. (b) Impermeable boundary instead; observation well 200 m south of the pumping well; $t=36$ h.

Find. Drawdown at each observation well/time using the method of images.

Approach. First compute the aquifer's $T=Kb$ and $S=S_sb$. A constant-head boundary is modelled by an image RECHARGE well (equal $-Q$) mirrored across the boundary, so the total drawdown is the real well's Theis drawdown minus the image well's; an impermeable (no-flow) boundary is modelled by an image PUMPING well (equal $+Q$) at the same mirror location, so the two drawdowns simply add. Place the pumping well at the origin with the boundary running north–south 200 m to the west; the image well sits a further 200 m west of the boundary, i.e. 400 m due west of the pumping well.

1. Aquifer properties. $$T=Kb=(1.2\times10^{-5})(30)=\boxed{3.6\times10^{-4}\ \text{m}^2/\text{s}},\qquad S=S_sb=(2.3\times10^{-5})(30)=\boxed{6.9\times10^{-4}}.$$
2. Part (a) — distances to the real and image wells. With the pumping well at $(0,0)$, the observation well is 60 m south, at $(0,-60)$; the image (recharge) well sits 400 m west, at $(-400,0)$: $$r_{\text{real}}=60\ \text{m},\qquad r_{\text{image}}=\sqrt{400^2+60^2}=\boxed{404.5\ \text{m}}.$$
3. Superposition at $t=24$ h. Computing $u=r^2S/4Tt$ and $W(u)$ separately for each well and subtracting (image well is a recharge well, so it partially offsets the real well's drawdown): $$s(24\text{h})=\frac{Q}{4\pi T}\Big[W(u_{\text{real}})-W(u_{\text{image}})\Big]=\boxed{9.52\ \text{m}}.$$
4. Superposition at $t=36$ h. Repeating with $t=129{,}600$ s (both $u$-values scale down, both $W$-values increase, but the real well's term still dominates): $$s(36\text{h})=\boxed{10.15\ \text{m}}.$$ Drawdown continues to grow between 24 and 36 hours (the river boundary limits, but does not fully arrest, the cone of depression's expansion at this distance).
5. Part (b) — impermeable boundary, obs. well 200 m south, $t=36$ h. Now $r_{\text{real}}=200$ m and $r_{\text{image}}=\sqrt{400^2+200^2}=447.2$ m, and because the image well is also a PUMPING well (no-flow boundary reflects rather than cancels), the two drawdowns ADD: $$s=\frac{Q}{4\pi T}\Big[W(u_{\text{real}})+W(u_{\text{image}})\Big]=\boxed{5.60\ \text{m}}.$$ Physically, the impermeable boundary forces all of the pumped water to be drawn from the aquifer on the well's own side, which is why the image well reinforces rather than offsets the real well's drawdown.